Mixing
![How to use dice to randomly select from 10 choices? [duplicate]](https://lh6.googleusercontent.com/-7Mng4mN1X2s/AAAAAAAAAAI/AAAAAAAAAAA/AAN31DW_XDkvv3xrhJACH5mIjz7MZXrO3Q/s72-c-mo/photo.jpg?sz=32)
Graphing absolute value inside absolute value equation
$begingroup$
How would I graph an equation with an absolute value inside an absolute value?
For example, $left|-left|xright|+1right|+left|y-2right|=3$
I tried graphing this on Desmos and it gave me some weird closed polygon. https://www.desmos.com/calculator/ysnqhkiuuk
graphing-functions absolute-value
asked Jan 13 at 5:22
ToylatteToylatte
173
$endgroup$
add a comment |
$begingroup$
How would I graph an equation with an absolute value inside an absolute value?
For example, $left|-left|xright|+1right|+left|y-2right|=3$
I tried graphing this on Desmos and it gave me some weird closed polygon. https://www.desmos.com/calculator/ysnqhkiuuk
graphing-functions absolute-value
asked Jan 13 at 5:22
ToylatteToylatte
173
$endgroup$
1
$begingroup$
wolframalpha.com/input/… This seems to validate what Desmos gave.
$endgroup$
– Moo
Jan 13 at 5:27
add a comment |
$begingroup$
How would I graph an equation with an absolute value inside an absolute value?
For example, $left|-left|xright|+1right|+left|y-2right|=3$
I tried graphing this on Desmos and it gave me some weird closed polygon. https://www.desmos.com/calculator/ysnqhkiuuk
graphing-functions absolute-value
asked Jan 13 at 5:22
ToylatteToylatte
173
$endgroup$
How would I graph an equation with an absolute value inside an absolute value?
For example, $left|-left|xright|+1right|+left|y-2right|=3$
I tried graphing this on Desmos and it gave me some weird closed polygon. https://www.desmos.com/calculator/ysnqhkiuuk
graphing-functions absolute-value
graphing-functions absolute-value
asked Jan 13 at 5:22
ToylatteToylatte
173
asked Jan 13 at 5:22
ToylatteToylatte
173
asked Jan 13 at 5:22
ToylatteToylatte
173
asked Jan 13 at 5:22
ToylatteToylatte
173
asked Jan 13 at 5:22
ToylatteToylatte
173
173
1
$begingroup$
wolframalpha.com/input/… This seems to validate what Desmos gave.
$endgroup$
– Moo
Jan 13 at 5:27
add a comment |
1
$begingroup$
wolframalpha.com/input/… This seems to validate what Desmos gave.
$endgroup$
– Moo
Jan 13 at 5:27
1
1
$begingroup$
wolframalpha.com/input/… This seems to validate what Desmos gave.
$endgroup$
– Moo
Jan 13 at 5:27
$begingroup$
wolframalpha.com/input/… This seems to validate what Desmos gave.
$endgroup$
– Moo
Jan 13 at 5:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You graph this equation manually the same way you graph any equation: try to find enough solution points to give you a feel for how the equation behaves, and interpolate the rest. When possible, use what you know about the functions to reduce the number of points needed.
In this case, you should know what $y=|x|$ looks like, which tells you that your equation is going to consist of line segments with vertices at places where the interior of the absolute values is $0$.
It is helpful to solve it for $y$:
$$|y - 2| = 3 - | 1 - |x|,|\y = 2pm(3 - | 1 - |x|,|)$$
Which also tells you that the graph will be symmetric about the line $y = 2$, and since $x$ only appears inside the absolute value, the graph will also be symmetric about $x = 0$.
If we examine just the quadrant where $x ge 0$ and $y ge 2$, we have
$$y = 5 - |1 - x|$$
For $x le 1, y = 5 - (1 -x) = 4 + x$. For $x ge 1, y = 5 + (1 - x) = 6 - x$ Note that since $y ge 2$ in this quadrant, this only holds when $6 - x ge 2$, thus $x le 4$. So in this quadrant, this is the graph of $$y = begin{cases} 4 + x& 0le x le 1\6 -x & 1 le x le 4end{cases}$$
Reflect that graph through $x = 0$, and then reflect again though $y = 2$ to get the full graph.
answered Jan 13 at 16:45
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071723%2fgraphing-absolute-value-inside-absolute-value-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You graph this equation manually the same way you graph any equation: try to find enough solution points to give you a feel for how the equation behaves, and interpolate the rest. When possible, use what you know about the functions to reduce the number of points needed.
In this case, you should know what $y=|x|$ looks like, which tells you that your equation is going to consist of line segments with vertices at places where the interior of the absolute values is $0$.
It is helpful to solve it for $y$:
$$|y - 2| = 3 - | 1 - |x|,|\y = 2pm(3 - | 1 - |x|,|)$$
Which also tells you that the graph will be symmetric about the line $y = 2$, and since $x$ only appears inside the absolute value, the graph will also be symmetric about $x = 0$.
If we examine just the quadrant where $x ge 0$ and $y ge 2$, we have
$$y = 5 - |1 - x|$$
For $x le 1, y = 5 - (1 -x) = 4 + x$. For $x ge 1, y = 5 + (1 - x) = 6 - x$ Note that since $y ge 2$ in this quadrant, this only holds when $6 - x ge 2$, thus $x le 4$. So in this quadrant, this is the graph of $$y = begin{cases} 4 + x& 0le x le 1\6 -x & 1 le x le 4end{cases}$$
Reflect that graph through $x = 0$, and then reflect again though $y = 2$ to get the full graph.
answered Jan 13 at 16:45
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
add a comment |
$begingroup$
You graph this equation manually the same way you graph any equation: try to find enough solution points to give you a feel for how the equation behaves, and interpolate the rest. When possible, use what you know about the functions to reduce the number of points needed.
In this case, you should know what $y=|x|$ looks like, which tells you that your equation is going to consist of line segments with vertices at places where the interior of the absolute values is $0$.
It is helpful to solve it for $y$:
$$|y - 2| = 3 - | 1 - |x|,|\y = 2pm(3 - | 1 - |x|,|)$$
Which also tells you that the graph will be symmetric about the line $y = 2$, and since $x$ only appears inside the absolute value, the graph will also be symmetric about $x = 0$.
If we examine just the quadrant where $x ge 0$ and $y ge 2$, we have
$$y = 5 - |1 - x|$$
For $x le 1, y = 5 - (1 -x) = 4 + x$. For $x ge 1, y = 5 + (1 - x) = 6 - x$ Note that since $y ge 2$ in this quadrant, this only holds when $6 - x ge 2$, thus $x le 4$. So in this quadrant, this is the graph of $$y = begin{cases} 4 + x& 0le x le 1\6 -x & 1 le x le 4end{cases}$$
Reflect that graph through $x = 0$, and then reflect again though $y = 2$ to get the full graph.
answered Jan 13 at 16:45
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
add a comment |
$begingroup$
You graph this equation manually the same way you graph any equation: try to find enough solution points to give you a feel for how the equation behaves, and interpolate the rest. When possible, use what you know about the functions to reduce the number of points needed.
In this case, you should know what $y=|x|$ looks like, which tells you that your equation is going to consist of line segments with vertices at places where the interior of the absolute values is $0$.
It is helpful to solve it for $y$:
$$|y - 2| = 3 - | 1 - |x|,|\y = 2pm(3 - | 1 - |x|,|)$$
Which also tells you that the graph will be symmetric about the line $y = 2$, and since $x$ only appears inside the absolute value, the graph will also be symmetric about $x = 0$.
If we examine just the quadrant where $x ge 0$ and $y ge 2$, we have
$$y = 5 - |1 - x|$$
For $x le 1, y = 5 - (1 -x) = 4 + x$. For $x ge 1, y = 5 + (1 - x) = 6 - x$ Note that since $y ge 2$ in this quadrant, this only holds when $6 - x ge 2$, thus $x le 4$. So in this quadrant, this is the graph of $$y = begin{cases} 4 + x& 0le x le 1\6 -x & 1 le x le 4end{cases}$$
Reflect that graph through $x = 0$, and then reflect again though $y = 2$ to get the full graph.
answered Jan 13 at 16:45
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
You graph this equation manually the same way you graph any equation: try to find enough solution points to give you a feel for how the equation behaves, and interpolate the rest. When possible, use what you know about the functions to reduce the number of points needed.
In this case, you should know what $y=|x|$ looks like, which tells you that your equation is going to consist of line segments with vertices at places where the interior of the absolute values is $0$.
It is helpful to solve it for $y$:
$$|y - 2| = 3 - | 1 - |x|,|\y = 2pm(3 - | 1 - |x|,|)$$
Which also tells you that the graph will be symmetric about the line $y = 2$, and since $x$ only appears inside the absolute value, the graph will also be symmetric about $x = 0$.
If we examine just the quadrant where $x ge 0$ and $y ge 2$, we have
$$y = 5 - |1 - x|$$
For $x le 1, y = 5 - (1 -x) = 4 + x$. For $x ge 1, y = 5 + (1 - x) = 6 - x$ Note that since $y ge 2$ in this quadrant, this only holds when $6 - x ge 2$, thus $x le 4$. So in this quadrant, this is the graph of $$y = begin{cases} 4 + x& 0le x le 1\6 -x & 1 le x le 4end{cases}$$
Reflect that graph through $x = 0$, and then reflect again though $y = 2$ to get the full graph.
answered Jan 13 at 16:45
Paul SinclairPaul Sinclair
19.7k21442
answered Jan 13 at 16:45
Paul SinclairPaul Sinclair
19.7k21442
answered Jan 13 at 16:45
Paul SinclairPaul Sinclair
19.7k21442
answered Jan 13 at 16:45
Paul SinclairPaul Sinclair
19.7k21442
19.7k21442
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071723%2fgraphing-absolute-value-inside-absolute-value-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
wolframalpha.com/input/… This seems to validate what Desmos gave.
$endgroup$
– Moo
Jan 13 at 5:27